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QJEEM23Ma9eed: JEE Main 2023 Mathematics - Class 12 JEEM23Ma9eed Solution: If the area bounded by the curve , lines and outside the circle …

If the area bounded by the curve 2y2=3x2 y^{2}=3 x , lines x+y=3,y=0x+y=3, y=0 and outside the circle (x−3)2+y2=2(x-3)^{2}+y^{2}=2 is A\mathrm{A} , then 4(π+4A)4(\pi+4 A) is equal to ____________.

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JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Area Under The Curves Question 44 English Explanation
amp;y2=3x2,x+y=3,y=0amp;2y2=3(3−y)amp;2y2+3y−9=0amp;2y2−3y+6y−9=0amp;(2y−3)(y+2)=0;y=3/2amp; Area (∫032(xR−x2)dy)−A1amp;=∫032((3−y)−2y23)dy−π8(2)amp;A=(3y−y22−2y39)032−π4amp;4 A+π=4[92−98−34]=212=10.50amp;∴4(4 A+π)=42\begin{aligned}& y^2=\frac{3 x}{2}, x+y=3, y=0 \\\\& 2 y^2=3(3-y) \\\\& 2 y^2+3 y-9=0 \\\\& 2 y^2-3 y+6 y-9=0 \\\\& (2 y-3)(y+2)=0 ; y=3 / 2 \\\\& \text { Area }\left(\int_0^{\frac{3}{2}}\left(x_R-x_2\right) d y\right)-A_1 \\\\& =\int_0^{\frac{3}{2}}\left((3-y)-\frac{2 y^2}{3}\right) d y-\frac{\pi}{8}(2) \\\\& A=\left(3 y-\frac{y^2}{2}-\frac{2 y^3}{9}\right)_0^{\frac{3}{2}}-\frac{\pi}{4} \\\\& 4 \mathrm{~A}+\pi=4\left[\frac{9}{2}-\frac{9}{8}-\frac{3}{4}\right]=\frac{21}{2}=10.50 \\\\& \therefore 4(4 \mathrm{~A}+\pi)=42\end{aligned}

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